Chemistry: The Central Science (13th Edition)

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.34a

Answer

The minimum frequency of light necessary to emit electrons from titanium is $1.047\times10^{15}s^{-1}$.

Work Step by Step

According to the formula: $$E=h\times \nu$$
Since $h$ is a constant, the minimum value of $E$ ($E_{min}$) would be achieved at the minimum value of $\nu$ ($\nu_{min}$). In other words, $$E_{min}=h\times\nu_{min}$$
From the question, we have:
- Minimum energy of photon for the metal to emit electrons: $E_{min}=6.94\times10^{-19}J$
- Also, the Planck's constant: $h\approx6.626\times10^{-34}J.s$
Therefore, the minimum frequency of light necessary to emit electrons from titanium is:
$\nu_{min}=\frac{E_{min}}{h}=\frac{6.94\times10^{-19}}{6.626\times10^{-34}}\approx1.047\times10^{15}s^{-1}$